本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。
本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。
batch-20261013-advance-20261022-r05-finite-group-foundations · 2026-10-13 · 100 題
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], 是)
結構基礎核對=(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], 是)
結構基礎核對=(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], 是)
結構基礎核對=(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}。請依 [n,table,closed] 欄序回答。
(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], 是)
結構基礎核對=(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], 是)
結構基礎核對=(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], 是)
結構基礎核對=(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], 是)
結構基礎核對=(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], 是)
結構基礎核對=(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], 是)
結構基礎核對=(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], 是)
結構基礎核對=(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}。請依 [n,table,closed] 欄序回答。
(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], 是)
結構基礎核對=(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], 是)
結構基礎核對=(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}。請依 [n,table,closed] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], 是)
結構基礎核對=(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], 是)
結構基礎核對=(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}。請依 [n,table,closed] 欄序回答。
(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], 是)
結構基礎核對=(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}。請依 [n,table,closed] 欄序回答。
(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], 是)
結構基礎核對=(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], 是)
結構基礎核對=(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}。請依 [n,table,closed] 欄序回答。
(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], 是)
結構基礎核對=(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}。請依 [n,table,closed] 欄序回答。
(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], 是)
結構基礎核對=(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-table-qualification；資料={"n":5,"table":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}。請依 [n,table,closed] 欄序回答。
(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], 是)
結構基礎核對=(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-table-qualification · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [], 是)
結構基礎核對=(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [], 是)
結構基礎核對=(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [], 是)
結構基礎核對=(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [], 是)
結構基礎核對=(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [], 是)
結構基礎核對=(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [], 是)
結構基礎核對=(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [], 是)
結構基礎核對=(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [], 是)
結構基礎核對=(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [], 是)
結構基礎核對=(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [], 是)
結構基礎核對=(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [], 是)
結構基礎核對=(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [], 是)
結構基礎核對=(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [], 是)
結構基礎核對=(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [], 是)
結構基礎核對=(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [], 是)
結構基礎核對=(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [], 是)
結構基礎核對=(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [], 是)
結構基礎核對=(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [], 是)
結構基礎核對=(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [], 是)
結構基礎核對=(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-associativity-audit；資料={"n":5,"table":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}。請依 [n,table,bad_triples,associative] 欄序回答。
(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [], 是)
結構基礎核對=(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-associativity-audit · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)
結構基礎核對=(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)
結構基礎核對=(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)
結構基礎核對=(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)
結構基礎核對=(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], 是)
結構基礎核對=(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], 是)
結構基礎核對=(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], 是)
結構基礎核對=(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], 是)
結構基礎核對=(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)
結構基礎核對=(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)
結構基礎核對=(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)
結構基礎核對=(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)
結構基礎核對=(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)
結構基礎核對=(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)
結構基礎核對=(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)
結構基礎核對=(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-identity-inverse-inventory；資料={"n":5,"table":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}。請依 [n,table,identities,inverse_rows,all_have_inverse] 欄序回答。
(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)
結構基礎核對=(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-identity-inverse-inventory · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], 是, 是, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], 是, 是, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], 是, 是, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)
結構基礎核對=(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], 是, 是, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], 是, 是, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)
結構基礎核對=(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], 是, 是, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], 是, 是, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)
結構基礎核對=(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], 是, 是, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)
結構基礎核對=(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], 是, 是, [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], 是, 是, [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], 是, 是, [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], 是)
結構基礎核對=(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], 是, 是, [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], 是, 是, [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], 是)
結構基礎核對=(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], 是, 是, [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], 是, 是, [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], 是)
結構基礎核對=(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], 是, 是, [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], 是)
結構基礎核對=(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], 是, 是, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], 是, 是, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], 是, 是, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], 是, 是, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], 是, 是, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], 是, 是, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], 是, 是, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)
結構基礎核對=(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], 是, 是, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)
結構基礎核對=(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], 是, 是, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], 是, 是, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], 是, 是, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], 是, 是, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], 是, 是, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)
結構基礎核對=(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], 是, 是, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], 是, 是, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)
結構基礎核對=(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], 是, 是, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)
結構基礎核對=(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], 是, 是, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)
結構基礎核對=(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], 是, 是, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], 是, 是, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)
結構基礎核對=(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], 是, 是, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], 是, 是, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)
結構基礎核對=(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], 是, 是, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], 是, 是, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)
結構基礎核對=(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], 是, 是, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate
完整有限Cayley表。操作=group-full-group-audit；資料={"n":5,"table":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}。請依 [n,table,closed,associative,identities,inverse_rows,group] 欄序回答。
(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)
結構基礎核對=(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], 是, 是, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], 是)；依完整明示資料重算；完整明示的精確切片不是一般定理或經驗聲明；結合律逐三元組核對；抽樣不證群公理。
有限群 · qf-str22-group-full-group-audit · candidate